A person walks at the rate of 3 km/hr. Rain appears to him in vertical direction at the rate of $3 \sqrt{3}$ km/hr. Find the magnitude and direction of true velocity of rain.

Moderate

A

6 km/hr, inclined at an angle of 45^o to the vertical towards the person's motion.
B

3 km/hr, inclined at an angle of 30o to the vertical towards the person's motion.
C

6 km/hr, inclined at an angle of 30^o to the vertical towards the person's motion.
D

6 km/hr, inclined at an angle of 60^o to the vertical towards the person's motion.

Solution

$\begin{array}{l} {\vec{v}}_{r / m} = {\vec{v}}_{r} - {\vec{v}}_{m} \\ {\vec{v}}_{r} = {\vec{v}}_{r / m} + {\vec{v}}_{m} \\ \begin{aligned} |{\vec{v}}_{r}| & = \sqrt{v_{r / m}^{2} + v_{m}^{2}} \\ = \sqrt{(3)^{2} + (3 \sqrt{3})^{2}} = 6 km / hr \\ \tan θ & = \frac{3}{3 \sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow θ = 30^{\circ} \end{aligned} \end{array}$

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